If the kinetic energy of the particle is increase to 16 times its previous value, the percentage change in the de Broglie wavelength of the paarticle is

Options

A. 0.25
B. 0.75
C. 0.6
D. 0.5

Answer

B

Explanation

Wavelength, `lamda = (h)/(p) = (h)/(sqrt(2mE)) or, lamda prop (1)/(sqrtE)`
`:. (lamda_(2))/(lamda_(1)) = sqrt((E_(1))/(E_(2))) = (1)/(4) or, lamda_(1) = 4lamda_(2)`
`:.` Percentage change in wavelength
`= (lamda_(1) - lamda_(2))/(lamda_(1)) xx 100 = (4lamda_(2) - lamda_(2))/(4lamda_(2)) xx 100 = 75`